A direct current of 10 A is superimposed on an alternating current / = 40 cos `omegat` (A) flowing through a wire. The effective value of the resulting current will be

Options

A. `10sqrt(2)` A
B. `20 sqrt(2)` A
C. `20 sqrt(3)` A
D. 30 A

Answer

To find the effective value of the resulting current when a direct current (DC) of 10 A is superimposed on an alternating current (AC) given by \( I = 40 \cos(\omega t) \) A, we can follow these steps:

### Step 1: Identify the Components of the Current
We have two components of current:
- Direct Current (DC): \( I_{DC} = 10 \, \text{A} \)
- Alternating Current (AC): \( I_{AC} = 40 \cos(\omega t) \, \text{A} \)

### Step 2: Calculate the RMS Value of the AC Current
The RMS (Root Mean Square) value of the AC current can be calculated using the formula:
\[ I_{rms} = \frac{I_{max}}{\sqrt{2}} \]
where \( I_{max} \) is the peak value of the AC current.

Here, the peak value \( I_{max} = 40 \, \text{A} \), so:
\[ I_{rms} = \frac{40}{\sqrt{2}} = \frac{40}{1.414} \approx 28.28 \, \text{A} \]

### Step 3: Calculate the Total Effective Current
The total effective current \( I_{total} \) when DC and AC are superimposed is given by:
\[ I_{total} = I_{DC} + I_{AC} \]
Since the DC component does not change with time, we can combine the RMS values:
\[ I_{rms, total}^2 = I_{DC}^2 + I_{rms, AC}^2 \]

### Step 4: Substitute the Values
Now we substitute the values we have:
\[ I_{rms, total}^2 = (10)^2 + (28.28)^2 \]
Calculating each term:
\[ I_{rms, total}^2 = 100 + 28.28^2 \approx 100 + 800 \approx 900 \]

### Step 5: Calculate the Final RMS Value
Now take the square root to find \( I_{rms, total} \):
\[ I_{rms, total} = \sqrt{900} = 30 \, \text{A} \]

### Conclusion
The effective value of the resulting current is:
\[ \boxed{30 \, \text{A}} \]

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